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63.30 Abstract trace formula

Suppose given an extension of profinite groups,

\[ 1 \to G \to \Gamma \xrightarrow {\deg } \widehat{\mathbf{Z}} \to 1 \]

We say $\Gamma $ has an abstract trace formula if and only if there exist

  1. an integer $q\geq 1$, and

  2. for every $d\geq 1$ a finite set $S_ d$ and for each $x\in S_ d$ a conjugacy class $F_ x \in \Gamma $ with $\deg (F_ x) = d$

such that the following hold

  1. for all $\ell $ not dividing $q$ have $\text{cd}_\ell (G)<\infty $, and

  2. for all finite rings $\Lambda $ with $q\in \Lambda ^*$, for all finite projective $\Lambda $-modules $M$ with continuous $\Gamma $-action, for all $n > 0$ we have

    \[ \sum \nolimits _{d|n}d \left( \sum \nolimits _{x \in S_ d} \text{Tr}( F_ x^{n/d} |_ M) \right) = q^ n \text{Tr}(F^ n|_{M \otimes _{\Lambda [[G]]}^{\mathbf{L}}\Lambda }) \]

    in $\Lambda ^\natural $.

Here $M \otimes _{\Lambda [[G]]}^{\mathbf{L}}\Lambda = LH_0(G, M)$ denotes derived homology, and $F=1$ in $\Gamma /G = \widehat{\mathbf{Z}}$.

Remark 63.30.1. Here are some observations concerning this notion.

  1. If modeling projective curves then we can use cohomology and we don't need factor $q^ n$.

  2. The only examples I know are $\Gamma = \pi _1(X, \overline\eta )$ where $X$ is smooth, geometrically irreducible and $K(\pi , 1)$ over finite field. In this case $q = (\# k)^{\dim X}$. Modulo the proposition, we proved this for curves in this course.

  3. Given the integer $q$ then the sets $S_ d$ are uniquely determined. (You can multiple $q$ by an integer $m$ and then replace $S_ d$ by $m^ d$ copies of $S_ d$ without changing the formula.)

Example 63.30.2. Fix an integer $q\geq 1$

\[ \begin{matrix} 1 & \to & G = \widehat{\mathbf{Z}}^{(q)} & \to & \Gamma & \to & \widehat{\mathbf{Z}} & \to & 1 \\ & & = \prod _{l\not\mid q} \mathbf{Z}_ l & & F & \mapsto & 1 \end{matrix} \]

with $FxF^{-1} = ux$, $u \in (\widehat{\mathbf{Z}}^{(q)})^*$. Just using the trivial modules $\mathbf{Z}/m\mathbf{Z}$ we see

\[ q^ n - (qu)^ n \equiv \sum \nolimits _{d|n} d\# S_ d \]

in $\mathbf{Z}/m\mathbf{Z}$ for all $(m, q)=1$ (up to $u \to u^{-1}$) this implies $qu = a\in \mathbf{Z}$ and $|a| < q$. The special case $a = 1$ does occur with

\[ \Gamma = \pi _1^ t(\mathbf{G}_{m, \mathbf{F}_ p}, \overline\eta ), \quad \# S_1 = q - 1, \quad \text{and}\quad \# S_2 = \frac{(q^2-1)-(q-1)}{2} \]


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